On Jun 12, 5:51*am, K1TTT wrote:
rather than worrying about basic physics, the real problem here is
that analysis of a non-linear circuit is being attempted using
techniques that are only valid for linear circuits.

Well, that's certainly one problem but I was talking about linear
circuit analysis. When will the effects of linear wave cancellation be
understood by the ham radio community?

Walt seems to have sidestepped the non-linear source problem by
defining the "source" as the first point in the system at which the
source signal becomes linear after filtering. He seems to draw a
system box boundary through that point and labels that signal, "the
source". The ratio of voltage to current at that point is, by
definition, the source impedance. That approach sure does simplify
things.

The output of a class-C amp is weighted with odd harmonics. The system
is obviously mismatched at the odd harmonic frequencies.
--
73, Cecil, w5dxp.com

On Jun 12, 1:23*pm, Cecil Moore wrote:
On Jun 12, 5:51*am, K1TTT wrote:

rather than worrying about basic physics, the real problem here is
that analysis of a non-linear circuit is being attempted using
techniques that are only valid for linear circuits.

Well, that's certainly one problem but I was talking about linear
circuit analysis. When will the effects of linear wave cancellation be
understood by the ham radio community?

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. they do superimpose which gives the illusion of
cancellation and intensification. the worst thing causing amateurs
problems is using power and trying to think about energy... neither
one is a linear phenomenon, but they are often treated as such because
of some simple cases where you can add and subtract powers... when
someone who doesn't understand the restrictions fully tries to use
powers on more general cases they get in trouble. that is why it is
always better to use current or voltage where you can (in many more
cases) add them linearly and be correct. of course then there are
those who try to push even those restrictions back into non-linear
realms, like amplifiers, and get into even more trouble.

On Jun 12, 9:00*am, K1TTT wrote:
they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

On Jun 12, 2:37*pm, Cecil Moore wrote:
On Jun 12, 9:00*am, K1TTT wrote:

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

On Jun 12, 2:37*pm, Cecil Moore wrote:
On Jun 12, 9:00*am, K1TTT wrote:

they won't because waves don't 'cancel' any more than they 'interact'
in linear systems. *they do superimpose which gives the illusion of
cancellation and intensification.

Here's proof otherwise:

b1 = s11*a1 + s12*a2 = 0 = reflected voltage toward the source

The permanent result is that reflected voltage and power equal zero?
How can that possibly be an illusion?

(b1)^2 = (s11*a1 + s12*a2)^2 = 0 = reflected power toward the source.
s11*a1 is not zero, s12*a2 is not zero, they are equal in magnitude
and opposite in phase.

What happens to the non-zero (s11*a1 + s12*a2)^2 power? The answer to
that question is what a lot of people are missing.

Those two reflected wavefronts have interacted destructively and
canceled. That destructive interference energy is redistributed back
toward the load as constructive interference energy in the other s-
parameter equation.

b2 = s21*a1 + s22*a2 = forward voltage toward the load

(b2)^2 = (s21*a1 + s22*a2)^2 = forward power toward the load

Doesn't the fact that reflected energy cannot be traced using only the
wave reflection model indicate that there is something missing from
that procedure?
--
73, Cecil, w6dxp.com

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

On Jun 12, 12:39*pm, K1TTT wrote:
i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

Too bad about the "s stuff". Here are the RF equations for a Z01 to
Z02 impedance discontinuity in a transmission line. The forward
voltage on the Z01 side is Vfor1 and the reflected voltage from the
impedance discontinuity (back toward the source) is Vref1. The forward
voltage on the Z02 side is Vfor2 and the reflected voltage (from the
load) is Vref2. Hopefully, the reflection and transmission
coefficients are self-explanatory. rho1 is the reflection coefficient
encountered by Vfor1, etc.

Vref1 = Vfor1(rho1) + Vref2(tau2) = 0

That is wavefront cancellation in action. The external reflection
phasor, Vfor1(rho1), is equal in magnitude and 180 degrees out of
phase with the internal reflection phasor, Vref2(tau2), arriving from
the mismatched load.

Vfor2 = Vfor1(tau1) + Vref2(rho2)

If these RF equations are normalized to SQRT(Z0), they are the same as
the s-parameter equations.
--
73, Cecil, w5dxp.com

On Jun 12, 9:17*pm, Cecil Moore wrote:
On Jun 12, 12:39*pm, K1TTT wrote:

i don't do s stuff so i have no idea what you just proved... give me
the impedances and voltages/currents.

Too bad about the "s stuff". Here are the RF equations for a Z01 to
Z02 impedance discontinuity in a transmission line. The forward
voltage on the Z01 side is Vfor1 and the reflected voltage from the
impedance discontinuity (back toward the source) is Vref1. The forward
voltage on the Z02 side is Vfor2 and the reflected voltage (from the
load) is Vref2. Hopefully, the reflection and transmission
coefficients are self-explanatory. rho1 is the reflection coefficient
encountered by Vfor1, etc.

Vref1 = Vfor1(rho1) + Vref2(tau2) = 0

That is wavefront cancellation in action. The external reflection
phasor, Vfor1(rho1), is equal in magnitude and 180 degrees out of
phase with the internal reflection phasor, Vref2(tau2), arriving from
the mismatched load.

Vfor2 = Vfor1(tau1) + Vref2(rho2)

If these RF equations are normalized to SQRT(Z0), they are the same as
the s-parameter equations.
--
73, Cecil, w5dxp.com

ok, so you defined a case where the superposition of the reflected and
refracted waves at a discontinuity results in a zero sum. is that
supposed to prove something? did i ever say that you could not define
such a case??

On Jun 12, 4:34*pm, K1TTT wrote:
ok, so you defined a case where the superposition of the reflected and
refracted waves at a discontinuity results in a zero sum. *is that
supposed to prove something? *did i ever say that you could not define
such a case??

I would call two waves superposing to zero indefinitely, "wave
cancellation". If that is not wave cancellation, where did the
reflected and refracted wavefronts go along with their energy
components? The answer to that question will reveal exactly what
happens to the reflected energy.

Here's a brain teaser for you and others. Given a Z01 to Z02 impedance
discontinuity with a power reflection coefficient of 0.25 at the '+'
discontinuity:

------Z01------+------Z02-------load

Pfor1 in the Z01 section is 100 watts. Pref1 in the Z01 section is
zero watts.

What is Pfor2, Pref2, and the SWR in the Z02 section?
--
73, Cecil, w5dxp.com